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#include <stdio.h>
#define sqr(x) (x*x)
int main()
{
int x=2;
printf("value of x=%d",sqr(x+1));
}
What is the value of x?
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Answer Posted / jugal
The OUTPUT of the program wud be
"value of x=9"
NOTE:
#define sqr(x) (x*x) and
#define sqr(x) x*x
are two very different things
what Divakar and all are saying is referring to the 2nd
one, where as in this case the 1st one is given
Coming to the actual question
The value of x will remain 2, since its value is not being
changed anywhere in the program, its just being passed to a
macro, but NOT modified there either.
Try adding a line at the end of the program
printf("x=%d",x);
| Is This Answer Correct ? | 1 Yes | 1 No |
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